This mechanism operates in all EM sources. It originates from the acceleration of electrons in Coulomb collisions with other electrons and with ions and nuclei. It comes from the German, 'brems' for braking, and 'strahlung' for radiation. The most common situation is the emission from a hot gas as the electrons collide with the nuclei, due to their random thermal motions. This is called 'thermal bremsstrahlung'. Bremsstrahlung can also occur when a beam of particles decelerates on encountering an obstacle. "Braking radiation" is the main way very fast charged particles lose energy when traveling through matter. Radiation is also emitted when charged particles are accelerated. In this case, the acceleration is caused by the electromagnetic fields of the atomic nuclei of the medium. These bremsstrahlung photons have a continuous spectrum with a broad peak of intensity, for photons with roughly half the incident electron energy, and are more numerous in directions perpendicular to the electrons' acceleration vector.
X-rays and shorter wavelengths can be easily generated by striking an anode with highly accelerated electrons within a high voltage gradient.
The higher the temperature of the cathode, the more electrons are released. We measure the level of current in milliamperes (mA). Increasing the current increases the number of electrons emitted from the cathode. This, in turn, increases the intensity of the rays produced. See the heated cathode lifter experiment.
Increasing the voltage (in kV), increases the speed of the electrons that strike the target. Higher potential difference settings produce shorter wavelength EM rays.
Doubling the tube current, doubles the quantity of heat produced. Heat production also varies almost directly with varying kVp. It is known that aluminium has got low Bremsstrahlung radiation levels, but there are other ways that it can emit short wavelength radiation.
The efficiency (not intensity) of x-ray & gamma ray production is independent of the hv current. Regardless of what current is selected, the efficiency of x-ray produciton remains constant. The efficiency of x-ray production increases with increasing projectile-electron energy. It may vary from 1% to 70% depending on the potential difference.
If the projectile electron interacts with an inner-shell electron of the target atom, rather than an outer-shell electron, characteristic x-radiation can be produced. Characteristic x-radiation results when the interaction is sufficiently violent to ionize the target atom, by the total removal of the inner-shell electron. Excitation of a inner (K)-shell electron does not immediately produce characteristic x-radiation.
When the projectile electron ionizes a target atom by removal of a K-shell electron, a temporary electron hole is produced in the K shell. This is a highly unnatural state for the target atom, and is corrected by an outer-shell electron falling into the hole in the K shell. The transition of an orbital electron from an outer shell to an inner shell is accompanied by the emission of an x-ray photon. Photons of this sort have energies that are, of course, characteristic of the anode material, and are emitted from the atom with equal probability in all directions. This x-ray has energy equal to the difference in the binding energies of the orbital electrons involved (K-L).
Example:
A K-shell elctron is removed from a tungsten atom and is repleced by an L shell electron. What is the energy of the characteristic x-ray that is emitted?
Answer:
For tungsten, K electrons have binding energies of 69.5 keV, and L electrons are bound by 12.1 keV. Therefore, the characteristic x-ray emitted has energy of: 69.5 - 12.1 = 57.4 keV
In summary, characteristic x-rays are produced by transitions of orbital electrons from outer to inner shells. Since the electron binding energy for every element is different, the characteristic x-rays produced in the various elements are also different. This type of x-radiation is called characteristic radiation, because it is characteristic of the target element. The effective energy characteristic x-rays increases with increasing atomic number of the target element.
When an electron hits the target so hard on its inner shell, the target material will radiate an EM wave of energy, equal to the energy difference between the innermost and the next shell called K & L1 respectively.
Luckily for us, the air molecules in the atmosphere make it impossible for any electron to travel the whole path from emitter to collector without any collision, in fact there would be millions of such collisions and the electrons can never obtain very high speeds, enough to generate any radiation upon impact. My latest tests for radiation within lifters confirmed no radiation when lifters are powered in air.
The following calculations apply only in pure vacuum.
For aluminium (our foil target), K = 1.56 keV , and L1 = 0.118 keV
The energy difference between the two shells is 1.442 keV.
So, when a charge of sufficient energy hits the aluminium foil, an energy packet of 1.442 keV will be released.
Using E = hf
1.442E3 x e/h = f
This gives f = 3.486E17 Hz, which happens to be the so called 'soft X-ray' band.
Alternatively, knowing the atomic number for Al (Z=13), approximate values can be found. Energy released by an electron shift from L to K = (Z-1)2 * 10.2 eV = 1.46 keV, and f=E/h gives 3.55E17 Hz.
As discussed, we have two kinds of radiation mechanisms radiating within the x-ray & gamma ray band. The resulting radiation will be similar to the above diagram. The characteristic radiation peaks will be added over a background noise floor, which is due to the Bremsstrahlung mechanism. The wavelengths of the characteristic lines are independent of the pd - they are just characteristic of the metal target. The background radiation cuts off sharply at A or B as the wavelengths diminish down to a certain point at which maximum radiation frequency occurs. This maximum limit for the radiation frequency will depend only upon the voltage across the electrodes, and is independent of the metal target characteristics.
hf(max) = eV
f(max)= eV/h
Conversely, if we need to know the minimum pd required to reach a particular maximum radiation frequency we use:
V(min)= f(max)h/e
So, suppose we need to reach the Gamma radiation band at 3E19 Hz, then
V(min) = 3E19 x 6.62e-34/ 1.6e-19
V(min) = 124 kV
Note that when operating pulsed dc supplies, sharp spikes on the secondary might reach higher hv values than the nominal output, and the duration of these spikes is significant at such frequencies.
